The triangles BOC and AOD are similar in two angles, since the angle BOC = AOD as vertical angles, the angle BOC = BDA as criss-crossing angles. The coefficient of similarity of triangles is: K = BC / AD = 5/10 = 1/2.
Then OB / OD = 1/2.
2 * BО = DO. DO + BO = 12 cm.
2 * BО + BО = 12 cm.
BO = 4 cm, BO = 8 cm.
Similarly, ОC = 3 cm, ОА = 6 cm.
In the triangle BOC, the Pythagorean theorem holds, BC ^ 2 = OB ^ 2 + OC ^ 2.
25 = 16 + 9.
Then the triangles BOC and AOD are rectangular.
Let’s draw the KН height through the point of intersection of the diagonals.
Then KO = ОВ * ОC / ВC = 4 * 3/5 = 12/5 = 2.4 cm.
HO = OA * OD / AD = 6 * 8/10 = 4.8 cm.
Then KН = KO + KН = 2.4 + 4.8 = 7.2 cm.
Determine the area of the trapezoid.
Savsd = (BC + AD) * KН / 2 = (5 + 10) * 7.2 / = 54 cm2.
Answer: The area of the trapezoid is 54 cm2.
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